I keep this question from my reading of Rindler's "Essential Relativity", where it's said, with no proof, that this metric accounts simply for the flat plane's but expressed in an unusual coordinate system (as is $ds^2=dx^2+x^2dy^2$ in polar coordinates). I tried to find the coordinate transformation seeking clues from known coordinate transformations. I've got not a lot of clues at all. From this point I've turned to the premises, trying to prove the fact of zero curvature. I've found it's not zero, but I am not sure about whether or not I've done it well. I've applied a formula from Wolfram mathworld Gaussian Curvature, this, the one that uses directly the metric tensor elements.

Has this metric curvature zero? If so, what is the coordinate transformation from cartesian to the coordinates that bring this line element?


This answer is just a more elementary (but not simpler), more calculatory and less geometric way to express what Ivo Terek and Ted Shifrin described.

Let $U \subset \mathbb{R}^2$ be an open set (which we don't exactly know for the moment) and let $\psi : U \to \mathbb{R}^2 : (x,y) \mapsto (u,v)$. Its derivative $D\psi$ is represented (in coordinates) by a matrix

$$ \left( \begin{array}{cc} \partial_x u & \partial_y u \\ \partial_x v & \partial_y v \end{array} \right) =: \left( \begin{array}{cc} A & B \\ C & D \end{array} \right) \, .$$

Assume that the Euclidean metric on the $(u,v)$-plane has the expression $y^2 dx^2 + x^2 dy^2$ with respect to the coordinates $(x,y)$. This means that $A^2+C^2 = y^2$, $B^2+D^2 = x^2$ and $AB+CD = 0$. By inspection, we see that setting $A= y \cos(f(x,y))$, $B=-x \sin(f(x,y))$, $C=y \sin(f(x,y))$ and $D = x \cos(f(x,y))$ with $f$ any function, we satisfy these three equations.

(Observe that this 'inspection' can be conceptually justified by first renormalizing the vectors $(D\psi) \partial_x$ and $(D \psi) \partial_y$ to be (ortho)normal as in Ivo Terek's answer and then performing an Euclidean rotation by an angle $f$ to the 'standard Cartesian orthonormal frame' as in Ted Shifrin's comment.)

Note however that $f$ has to be chosen in such a way that $A = \partial_x u$, etc. If such a choice exists, then since $\partial_y \partial_x u = \partial_x \partial_y u$, we have the constraint $\partial_y A = \partial_x B$, that is

$$ \cos f - f'_y y \sin f = - \sin f - f'_x x \cos f \, \mbox{ i.e } \, (1- f'_y y) \sin f = -(1 + f'_x x) \cos f. $$

With a similar reasoning for $v$, we have the constraint $\partial_y C = \partial_x D$ which yields

$$ \sin f + f'_y y \cos f = \cos f - f'_x x \sin f \, \mbox{ i.e } \, (1 + f'_x x ) \sin f = (1 - f'_y y ) \cos f. $$

This is possible only if $-f'_x x = 1 = f'_y y$, whence $$f'_x = -1/x \Rightarrow f = - \log x + g(y) \Rightarrow f'_y = g'(y) = 1/y \Rightarrow g(y) = \log y + C \, .$$

We thus obtain $f(x,y) = \log(y/x) + C$ for some constant $C$, which we shall choose to be $0$ for simplicity. We now know what should be the functions $A, \dots, D$ if the function $\psi$ is to exist (because of the Poincaré lemma, if $U$ is simply-connected, such a map $\psi$ exists ; note also that for $A, \dots, D$ to be defined, the coordinates $x$ and $y$ can't vanish on $U$).

What remains to be done is to compute $u$ and $v$...

Observe that setting $U = -xy \sin \log(y/x)$ and $V = xy \cos \log(y/x)$, we get $\partial_x U = -C + A$, $\partial_y U = B -D$, $\partial_x V = A + C$ and $\partial_y V = D + B$. It is easy from there to check that $u = (U+V)/2$ and $v = (V-U)/2$ are solutions to the problem.

  • $\begingroup$ Cear and nice. I need now to familarize with the more geometric treatment, as Ted Shifrin suggest, to get insight on the, say, preliminaries. $\endgroup$ – Rafa Budría Feb 5 '17 at 11:56

I'll give a partial answer. Let's compute the curvature. We have a an orthonormal frame $$E_1 = \frac{1}{y}\partial_x, \quad E_2 = \frac{1}{x}\partial_y,$$with dual coframe given by $$\theta_1 = y\,{\rm d}x, \quad \theta_2 = x\,{\rm d}y.$$We also have $${\rm d}\theta_1 = -{\rm d}x \wedge {\rm d}y, \quad {\rm d}\theta_2 = {\rm d}x\wedge {\rm d}y$$Now we want to find the connection form $$\omega_{12} = A\,{\rm d}x+B\,{\rm d}y$$ satisfying $${\rm d}\theta_1 = \omega_{12}\wedge \theta_2 \quad\mbox{and}\quad {\rm d}\theta_2 = -\omega_{12}\wedge \theta_1.$$The first relation gives $$-{\rm d}x\wedge {\rm d}y = Ax\,{\rm d}x\wedge {\rm d}y \implies A = -\frac{1}{x}$$and the second gives $${\rm d}x \wedge {\rm d}y = -By\,{\rm d}y\wedge {\rm d}x \implies B = \frac{1}{y}.$$Now we use theorema Egregium: ${\rm d}\omega_{12} = -K \theta_1 \wedge \theta_2$: $$\begin{align*} {\rm d}\omega_{12} &= {\rm d}\bigg( -\frac{1}{x}\,{\rm d}x + \frac{1}{y} \,{\rm d}y \bigg)\end{align*} = 0,$$and hence $K=0$.

I'm thinking for a bit already but I didn't saw the coordinate change needed to put the metric in the form ${\rm d}s^2 = {\rm d}u^2+{\rm d}v^2$ yet.

  • $\begingroup$ Thank you. I make the first question as answered, but I'd be grateful if you give one to the second. I think the answer cannot be easy even to express; maybe at the best an implicit equation... No idea really. $\endgroup$ – Rafa Budría Feb 4 '17 at 19:52
  • $\begingroup$ One idea is to solve the geodesic equations for that metric - we know that the equations that'll show up will represent straight lines on the plane, and then use this to define suitable coordinates. I'll think more about it... $\endgroup$ – Ivo Terek Feb 4 '17 at 20:17
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    $\begingroup$ I did. I have the possibility to click up, indeed :) $\endgroup$ – Rafa Budría Feb 4 '17 at 21:55
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    $\begingroup$ Rotate the given frame field by angle $\theta = \log(x/y)$. Then in the new frame field we'll have $\tilde\omega_{12}=0$, which means that $d\tilde\theta_1 = d\tilde\theta_2 = 0$, so $\theta_i = du_i$ locally. This means that we'll have $\tilde\theta_1 = \cos(\log x/y)(y\,dx) + \sin(\log x/y)(x\,dy)$. This form is in fact closed, therefore exact on $xy\ne 0$. One can find the potential by the usual algorithms, but it is far from pretty. Similarly for $\tilde\theta_2$. $\endgroup$ – Ted Shifrin Feb 4 '17 at 23:17
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    $\begingroup$ P.S. If you're not too familiar with the moving frame set-up, I recommend a very succinct treatment in my differential geometry text. The idea of my solution here appears in Exercise 3.3.13. [This is addressed to @RafaBudría, not Ivo.] $\endgroup$ – Ted Shifrin Feb 4 '17 at 23:56

Just happened to fall over this post. Don't know if you are still interested in getting simple methods for obtaining this metric?

Anyway, writing $$ ds^2 = x^2y^2 \left( \frac{dx^2}{x^2} + \frac{dy^2}{y^2} \right) $$ suggests at first to set $(x,y)=(e^u,e^v)$ to obtain a conformal metric: $$ ds^2 = e^{2u + 2v} (du^2 + dv^2) $$ From here we may use (complex) conformal methods to do the manipulations. Writing $z=u+iv$ note that $2u+2 v = 2 {\rm Re} ((1-i)(u+iv)) = 2 {\rm Re} ((1-i)z) $ so the metric may be written $$ ds^2 = (f(z) \overline{f(z)}) \; dz \overline {dz} = |f(z)dz|^2$$ with $f(z)=e^{(1-i)z}$ which is holomorphic in $z$. Without further computation you may then conclude that the metric is flat, since it has the form $ds^2 = |dw|^2$ where $w$ is a primitive of $f(z)dz$.

For explicit coordinates you may choose (multiplying by a constant of modulus one doesn't change the metric): $$ w = \frac{1}{|1-i|} e^{(1-i)z} = \frac{1}{\sqrt{2}}e^{u+v} e^{i(v-u)} = \frac{1}{\sqrt{2}} xy (\cos \ln (y/x) + i \sin \ln (y/x)) $$ So $ds^2 = da^2 + db^2$ with $a=\frac{1}{\sqrt{2}}xy \;\cos \ln (y/x)$ and $b=\frac{1}{\sqrt{2}} xy \;\sin \ln (y/x)$

  • $\begingroup$ Yes, I am still interested :) Thank you very much. $\endgroup$ – Rafa Budría Sep 20 '17 at 5:06

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